Show that any atlas of a compact $C^\alpha$-manifold has at least two elements

Question

Basically I'm tasked with the following:

Let K $\subset \mathbb{R}^n$ be a compact $C^\alpha$-manifold. Show that an atlas of K consists of at least two charts.

I've tried using contradiction: Assume that there exists some chart $\Phi$ and some open sets U $\subset \mathbb{R}^n$ with K $\subset U$ and V $\subset \mathbb{R}^k$ such that $$\Phi \in C^\alpha(V,\mathbb{R}^n) \:\wedge \forall x \in V: rk(D\Phi(x)) = k$$

But now I'm struggling to find the contradiction: I know that K $\subset$ U and there exists some finite open cover $\{ U_i \}$ of K. But then again $\Phi$ doesn't necessarily have to be defined on this open covering and I don't see how this leads anywhere.

I also know that $\Phi^{-1}(K)$ needs to be compact as a image of a compact set under a continous map and I tried constructing some contradiction using the image of the boundary and the interior but that didn't work out either.

Has anyone got a tip?

Answer 1

We have to assume $K \ne \emptyset$ and $n > 0$.

An atlas of $K$ covers $K$ by open subsets of $K$ which are homeomorphic to open subsets of $\mathbb R^n$. If there exists an atlas with a single chart, then $K$ itself is homeomorphic to a open subset $V$ of $\mathbb R^n$. Since we assumed $K \ne \emptyset$, we must have $V \ne \emptyset$.

This is impossible because $K$ is compact and nonempty open subsets of $\mathbb R^n$ are not compact if $n > 0$.

Answer 2

Okay, I just realized how trivial this is:

U has to be an open subset of K, therefore $U \subset K$. Furthermore since $\{ \Phi: V \rightarrow U \}$ is an atlas von K, we have that $K \subset U$, therefore $$K = U$$.

Now $\Phi$ is a homeomorphism, therefore $V = \Phi^{-1}(K)$ is compact. Furthermore V is open by definition. This is a direct contradiction, since V is now open and closed and therefore either $V = \mathbb{R}^k$ or $V = \emptyset$. However V can't be equal to $\mathbb{R}^k$ since it's compact and $\mathbb{R}^k$ is only compact for k = 0. If k = 0, then $\Phi$ would be a constant map since $rk(D\Phi) = 0$ and then U = $\Phi(V)$ wouldn't be open. V also can`t be equal to $\mathbb{R}$ since K is non-empty, therefore it contradicts our inital assumption.